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3 years ago

When is balance achieved between the forward and reverse reactions?

1 answer:

8 0

Equilibrium occurs when forward and reverse directions of a reversible reaction occur at the same rate so there is no overall change in the amounts of reactants and products.

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Does the bowling ball have more potential energy or kinetic energy as it is halfway through its fall? Why?

ahrayia [7]

Answer:

more kinetic

Explanation:

I think the kinetic energy is 75 percent while the potential energy is 25 percent

5 0

3 years ago

Metric conversions. Please help ASAP.

lesya692 [45]

Answer:

14. 13.2cg = 1.32dg

15. 3.8m = 0.0038km

16. 24.8L = 24800mL

17. 0.87kL = 870L

18. 26.01cm = 0.0002601km

19. 0.001hm = 10cm

Explanation:

14. 13.2/10 = 1.32

15. 38/1000 = 0.0038

16. 24.8(1000) = 24,800

17. 0.87(1000) = 870

18. 26.01/100000 = 0.0002601

19. 0.001hm(10000) = 10

An easy way to do these by yourself is to familiarize yourself with what each prefix means. Once you do this, you can multiply the value of the prefix when converting from a smaller unit of measurement to a larger one and divide the value of the prefix when converting from a large unit of measurement to a smaller one.

7 0

3 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

Which of the following correctly identifies two of the eight elements of weather

Shtirlitz [24]

Are there any options?

8 0

3 years ago

How do you convert kilograms to pounds?

Maru [420]

Answer: (1 Kilogram = 2.20462 pounds) . There are 2.2046226218 lb in 1 kilogram. To convert kilograms to pounds, multiply your figure by 2.205 for an approximate result. 1 kilogram is also equal to 2 lb and 3.27396195 oz. Working out a rough estimate in your head for converting to pounds and ounces may be tricky - remember that there are 16 ounces in a pound.

4 0

2 years ago